Comment 2 for bug 1087423

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Davanum Srinivas (DIMS) (dims-v) wrote :

There is a run_server in wsgi.py is that enough? If not, please reopen this bug.

def run_server(application, port):
    """Run a WSGI server with the given application."""
    sock = eventlet.listen(('0.0.0.0', port))
    eventlet.wsgi.server(sock, application)